How to Simplify (3.25) using Rodrigues Formula and Evaluate the Derivative?

[Bill Foster]

Homework Statement

How do you get from (3.25) to (3.26) in Jackson?

Homework Equations

Equation 3.25:

<br /> A_l=\left(2l+1\right)\int_0^1P_l\left(x\right)dx<br />

Equation 3.26:

<br /> A_l=\left(-\frac{1}{2}\right)^{\frac{l-1}{2}}\frac{\left(2l+1\right)\left(l-2\right)!}{2\left(\frac{l+1}{2}\right)!}<br />

Rodriques:

<br /> P_l\left(x\right)=\frac{1}{2^l l!}\frac{d^l}{dx^l}\left(x^2-1\right)^l<br />

The Attempt at a Solution

Put Rodrigues into (3.25):

<br /> A_l=\frac{\left(2l+1\right)}{2^l l!}\int_0^1\frac{d^l}{dx^l}\left(x^2-1\right)^ldx<br />

Evaluate:

<br /> \frac{d^l}{dx^l}\left(x^2-1\right)^l<br />

<br /> =\frac{d^l}{dx^l}\sum_k{\left(\frac{l!}{k!\left(l-k\right)!}\right)\left(x^2\right)^{l-k}\left(-1\right)^k}<br />

We worked this out in class, and somehow that sum goes away. But I forgot how.

As I work out the derivatives, I end up with:

<br /> =\sum_k{\left(\frac{l!}{k!\left(l-k\right)!}\right)2\left(l-k\right)\left(2\left(l-k\right)-1\right)\left(2\left(l-k\right)-2\right)\left(2\left(l-k\right)-3\right)...\left(2\left(l-k\right)-\left(l-1\right)\right)x^{l-2k}\left(-1\right)^k<br />

Any value of k&gt;\frac{l}{2} and the expression is zero. So the sum only goes from k=0 to \frac{l-1}{2}.

How do I get rid of that sum completely?

 

[gabbagabbahey]

Equation 3.26:

<br /> A_l=\left(-\frac{1}{2}\right)^{\frac{l-1}{2}}\frac{\left(2l+1\right)\left(l-2\right)!}{2\left(\frac{l+1}{2}\right)!}<br />

Looks like Jackson contains a typo; this should should be:

<br /> A_l=(-1)^{l-1}\left(\frac{1}{2}\right)^{\frac{l-1}{2}}\frac{\left(2l+1\right)\left(l-2\right)!}{2\left(\frac{l+1}{2}\right)!}<br />

Evaluate:

<br /> \frac{d^l}{dx^l}\left(x^2-1\right)^l<br />

<br /> =\frac{d^l}{dx^l}\sum_k{\left(\frac{l!}{k!\left(l-k\right)!}\right)\left(x^2\right)^{l-k}\left(-1\right)^k}<br />

We worked this out in class, and somehow that sum goes away. But I forgot how.

You've messed up the binomial expansion; you should have:

<br /> =\frac{d^l}{dx^l}\sum_{k=0}^{\infty}{\left(\frac{l!}{k!\left(l-k\right)!}\right)x^{2k}\left(-1\right)^{l-k}}<br />

 

[Bill Foster]

That still doesn't get rid of the sum. It only gets rid of the values k=0 to k=\frac{l-1}{2}.

 

[gabbagabbahey]

Okay, so that leaves you with

A_l=\frac{2l+1}{2^ll!}\sum_{k=\frac{l-1}{2}}^\infty\frac{l!}{k!(l-k)!}}(-1)^{l-k}\int_0^1\frac{d^l}{dx^l}(x^{2k})dx

Now, I think what you want to do is make the substitution j=k-\frac{l-1}{2}...

 

[Bill Foster]

\frac{d^l}{dx^l}\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)x^{2k}\left(-1\right)^{l-k}}

Differentiate once (1st time):

\frac{d^{l-1}}{dx^{l-1}}\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2kx^{2k-1}\left(-1\right)^{l-k}}

Differentiate again (2nd time):

\frac{d^{l-2}}{dx^{l-2}}\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)x^{2k-2}\left(-1\right)^{l-k}}

Differentiate again (third time):

\frac{d^{l-3}}{dx^{l-3}}\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)x^{2k-3}\left(-1\right)^{l-k}}

...
...

Differentiate lth time:

\frac{d^{l-l}}{dx^{l-l}}\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}

That leaves me with:

\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}

Now, if k=0, the whole expression is zero. So we can rewrite it as:

\sum_{k=1}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}

But if k=\frac{1}{2}, then the expression is also zero. But k can only be an integer. However, due to the term 2k-2, if k=1 the expression is zero. So rewrite as:

\sum_{k=2}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}

Now let's just cut to the chase:

(2k-(l-1)

If k=\frac{l-1}{2} or less, the expression is zero.

So rewrite as:

\sum_{k=\frac{l-1}{2}+1}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}

Which can be written as:

\sum_{k=\frac{l+1}{2}}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}

That only works if l is odd. If l is even, then it would be written as:

\sum_{k=\frac{l}{2}}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}

So we have the following:

A_l=\frac{\left(2l+1\right)}{2^l l!}\int_0^1\frac{d^l}{dx^l}\left(x^2-1\right)^ldx
=\frac{\left(2l+1\right)}{2^l l!}\int_0^1\sum_{k=\frac{l}{2}}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}dx
=\frac{\left(2l+1\right)}{2^l l!}\sum_{k=\frac{l}{2}}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))\left(-1\right)^{l-k}\int_0^1x^{2k-l}}dx for even l
=\frac{\left(2l+1\right)}{2^l l!}\sum_{k=\frac{l+1}{2}}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))\left(-1\right)^{l-k}\int_0^1x^{2k-l}}dx for odd l

Now: \int_0^1x^{2k-l}dx=\frac{1}{2k-l+1}x^{2k-l+1} evaluated from 1 to 0:

That would equal:

\int_0^1x^{2k-l}dx=\frac{1}{2k-l+1}

So now we have:

A_l=\frac{\left(2l+1\right)}{2^l l!}\sum_{k=\frac{l}{2},\frac{l+1}{2}}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))\left(-1\right)^{l-k}\frac{1}{2k-l+1}

Which can be written as:

A_l=\frac{\left(2l+1\right)}{2^l l!}\sum_{k=\frac{l}{2},\frac{l+1}{2}}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-2))\left(-1\right)^{l-k}

Now I still have that sum.

 

[gabbagabbahey]

You might want to re-read my last post; summing over j instead of k will probably make things easier for you...

 

[Bill Foster]

You might want to re-read my last post; summing over j instead of k will probably make things easier for you...

I don't see how. It changes the limits on the sum:

A_l=\frac{\left(2l+1\right)}{2^l l!}\sum_{j=1}^{\frac{l+1}{2}}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-2))\left(-1\right)^{l-k}

And all the k's become more complicated:

A_l=\frac{\left(2l+1\right)}{2^l l!}\sum_{j=1}^{\frac{l+1}{2}}{\left(\frac{l !}{\left(j+\frac{l-1}{2}\right)!\left(l-\left(j+\frac{l-1}{2}\right)\right)!}\right)2\left(j+\frac{l-1}{2}\right)(2\left(j+\frac{l-1}{2}\right)-1)(2\left(j+\frac{l-1}{2}\right)-2)...(2\left(j+\frac{l-1}{2}\right)-(l-2))\left(-1\right)^{l-\left(j+\frac{l-1}{2}\right)}

 

[gabbagabbahey]

Why are you summing from j=1 to j=(l+1)/2?

You started out with something like

A_l=\sum_{k=0}^{\infty}c_k\int_0^1\frac{d^l}{dx^l}x^{2k}dx=\sum_{k=0}^{\infty}c_k\left.\left[\frac{d^{l-1}}{dx^{l-1}}x^{2k}\right]\right|_0^1

The terms for which 2k&lt;l-1 are all zero, and l is odd (Remember, Jackson showed the A_l=0 for even l, so this sum is just to determine the odd coefficients), So the first \frac{l-1}{2}-1 terms of the sum are zero and you are left with:

A_l=\sum_{k=\frac{l-1}{2}}^{\infty}c_k\left.\left[\frac{d^{l-1}}{dx^{l-1}}x^{2k}\right]\right|_0^1

So the substitution j=k-\frac{l-1}{2}should give you

A_l=\sum_{j=0}^{\infty}c_{j+\frac{l-1}{2}}\left.\left[\frac{d^{l-1}}{dx^{l-1}}x^{2j+l-1}\right]\right|_0^1=\sum_{j=0}^{\infty}c_{j+\frac{l-1}{2}}\left.\left[(2j+l-1)(2j+l-2)\ldots(2j+1)x^{2j}\right]\right|_0^1=\sum_{j=0}^{\infty}c_{j+\frac{l-1}{2}}\frac{(2j+l-1)!}{(2j)!}

 

[Bill Foster]

The sum doesn't go to infinity. Using k as the index variable, it goes from 0 to l:

<br /> \frac{d^l}{dx^l}\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)x^{2k}\left(-1\right)^{l-k}}<br />

 

[gabbagabbahey]

No, it goes to infinity. You need to review the binomial expansion.

(x^2-1)^l=\sum_{k=0}^{\infty}{\left(\frac{l !}{k!\left(l-k\right)!}\right)x^{2k}\left(-1\right)^{l-k}}\neq\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)x^{2k}\left(-1\right)^{l-k}}

 

[Bill Foster]

No, it goes to infinity. You need to review the binomial expansion.

(x^2-1)^l=\sum_{k=0}^{\infty}{\left(\frac{l !}{k!\left(l-k\right)!}\right)x^{2k}\left(-1\right)^{l-k}}\neq\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)x^{2k}\left(-1\right)^{l-k}}

http://en.wikipedia.org/wiki/Binomial_theorem

 

[Bill Foster]

Why are you summing from j=1 to j=(l+1)/2?

Because if k=\frac{l-1}{2}, that term is zero. So the sum has to start at k=\frac{l+1}{2}.

If j=k-\frac{l-1}{2}, then the first term for j is when k=\frac{l+1}{2}:

j=k-\frac{l-1}{2} = \frac{l+1}{2}-\frac{l-1}{2} = 1.

The last term is k = l

So j=k-\frac{l-1}{2} = l-\frac{l-1}{2} = \frac{l+1}{2}

 

[gabbagabbahey]

http://en.wikipedia.org/wiki/Binomial_theorem

Contrary to popular belief, wikipedia is not always the most accurate source for information.

Try this link instead. Or, better yet, break out your old calculus textbook.

Edit: After my morning cup of coffee, I realized that there is nothing wrong with only summing up to k=l. However, the sum can be extended to infinity since \begin{pmatrix}l\\k\end{pmatrix} is zero for k&gt;l.

 

[Bill Foster]

Contrary to popular belief, wikipedia is not always the most accurate source for information.

Try this link instead. Or, better yet, break out your old calculus textbook.

I would like to direct your attention to equation 2 in that link, and the text that immediately precedes it.

Edit: After my morning cup of coffee, I realized that there is nothing wrong with only summing up to k=l. However, the sum can be extended to infinity since \begin{pmatrix}l\\k\end{pmatrix} is zero for k&gt;l.

It would only go to infinity if l goes to infinity.

 

[gabbagabbahey]

Anyways, can you agree that

A_l=-\frac{2l+1}{2^l}\sum_{j=0}^{j=\frac{l+1}{2}}\frac{(-1)^j(2j+l-1)!}{(j+\frac{l-1}{2})!(\frac{l+1}{2}-j)!(2j)!}

?

 

[Bill Foster]

Anyways, can you agree that

A_l=-\frac{2l+1}{2^l}\sum_{j=0}^{j=\frac{l+1}{2}}\frac{(-1)^j(2j+l-1)!}{(j+\frac{l-1}{2})!(\frac{l+1}{2}-j)!(2j)!}

?

It looks clearer written in terms of k.

 

[gabbagabbahey]

Since l is odd, make the substitution l=2n+1...